Linearly smoothed Möbius cancellation from finite zero-free rectangles (${\sum }_{n\le x}\mu (n)(x-n)=o(x^2)$)

The Perron's formula kernel $y^{s+1}/(s(s+1))$ integrates to $(y-1{)}_{+}$, giving the smoothed sum as an integral of $x^{s+1}/(\zeta (s)s(s+1))$. For each fixed height $T$, a zero-free rectangle permits shifting the central contour to a real part below one; its vertical contribution is $o(x^2)$, and its horizontal contributions are $O_T(x^2/\log x)$ when the original real part is $1+1/\log x$. A bound $1/\zeta (\sigma +it)=O({\log }^7(2+|t|))$ for $\sigma \ge 1$ bounds the remaining tails by $C{x}^2{\int }_T^{\infty }{\log }^7(2+t)t^{-2}dt$. Letting $x$ grow first and $T$ afterward proves the displayed cancellation, without uniform estimates to the left of one.